HYDROSTATIC LUBRICATION
Circular Footstep Bearing
Load W
Flow Out
at p = 0
Flow Out
Circular Recess
(p = p0)
The lubricant(oil) is pumped at a high pressure to the pocket
(circular recess). Oil flows radially outward through the
narrow gap. The pressure distribution (which varies from
pumping pressure at the recess to atmospheric pressure) on
the shaft lift the load W.
Flow In
at p = p0
Preliminary Theory (Slot Leakage formula)
Consider steady viscous flow between very close
parallel plates, under the pressure gradient dp/dx.
It can be shown that the velocity and shear stress
distribution are given by u = (1/2)(-dp/dx)[(t/2)2 - y2] and
The discharge per unit width
= (dp/dx)y
q = (t3/12)(-dp/dx). This is known as slot leakage formula.
r2
This formula can be used to analyse the footstep bearing. Consider a shaft of
outer radius r2 located over a plane pad with a circular recess of radius r1. The
chamber is supplied with oil at pressure p0. The radial flow rate through the
section at radius r is
Q = q(2r) = (t3/6)(-dp/dx) r.
Q
r1
p0
For the continuity Q should be a constant (ie. independent of r). Then by integrating this equation, the
pressure distribution can be obtained, with the boundary condition at r = r2 p = 0, as
r1
r2
p = (6Q/t3)loge[r2/r] ;
r1 r r2 .
At r = r1 , p = p0. Therefore p0 = (6Q/t3)loge[r2/r1]
Then
p = p0loge[r2/r]/loge[r2/r1]
2
The load carrying capacity W =pdA =p(2rdr) = p0(r2 - r1 )/{2loge[r2/r1]} = (3Q/t
)(r22
p0
- r12).
HYDRODYNAMIC LUBRICATION
It is known that a journal bearing operates efficiently when it is brought up to a certain speed. This is due to
hydrodynamic action. A film of oil which completely separates the surfaces is maintained by relative motion
of the two surfaces. The thrust bearing also operates on the same principle.
(a) Thrust Bearings
Tilted Pad Bearing
The simplest trust bearing consists of a slider and a bearing pad,
which is slightly inclined to it.
Slide
r
Bearing Pad
The flow rate per unit width across the section x-x is
Q = Ut/2 + (t3/12)(-dp/dx) .
Since p = 0 at x = b/2 , there must be a point - b/2 < x < b/2 where the pressure is maximum (ie. dp/dx = 0).
Let at t = t0 , x = x0 and dp/dx = 0. Then Q = Ut0/2 and for the continuity Ut0/2=Ut/2+(t3/12)(-dp/dx).
Therefore dp/dx = 6U(t - t0)/t3. This is known as Reynolds Equation in one dimension.
Now using the relation (h - t)/x = 2e/b and the boundary conditions at t = h e, p = 0; the above equation can
be integrated to show that t0 = (h2 - e2)/h ,
x0 = eb/2h ,
p =
3 Ub 2
3 Ube b 2 4 x 2
2
e (h t ) =
2he
2h b 2 h 2 4e 2 x 2
and
b/2
p max = [ p ]t = t0 =
The load carrying capacity per unit width of the pad isW = -b/2 p dx =
3 Ube
.
2h(h2 - e2 )
3 Ub2
h + e 2e
log e
- .
2
2e
h-e h
The frictional force acting on the slider per unit width (drag force) is given by
du
2 Ub
h + e 3e
b/2
b/2
F = -b/2 dx = -b/2 dx =
loge
- .
e
h - e 2h
dy
If / is the virtual coefficient of friction for the slider then
' =
4e
F
h + e 3e
h + e 2e
=
- log e
- .
log e
W
3b
h - e 2h
h - e h
The frictional force acting on the bearing pad F0 = Fcos - Wsin ; where = tan-1[2e/b] is the slope of the
pad.
If / = tan;
F0 = Wcos[tan - tan].
Therefore it is clear that when > F0 becomes negative. That is the direction of flow of fluid near the pad is
reversed. In order to prevent this situation . Therefore the critical slope *=. By using the expressions
for tan=/ and tan, it can be shown that at the critical slope e/h = 0.8586. Thus for a given h and b
1.7172(h/b).
Rayleigh Step Bearing
The flow rate per unit width acrossany cross section is given by
Slider
Q = Ut/2 + (t /12)(-dp/dx).
Then for the continuity
= Ut1/2 + (t13/12)(-dp/dx)1
= Ut2/2 + (t23/12)(-dp/dx)2 = constant.
Bearing Pad
Therefore (-dp/dx)1 = constant and (-dp/dx)2 = constant. Let the maximum pressure is po. Then (dp/dx)1 =
po/b1 and (dp/dx)2 = -po/b1. Substitution of these expressions in the expressions for the flow rate leads to
po = 6Ub1b2(t1 - t2)/(b2t13 + b1t23).
The load carrying capacity per unit width of the pad is
W = (0b1+b2 ) p dx =
1
3 U( b1 + b2 ) b1 b2 ( t 1 - t 2 )
p0 (t 1 + t 2 ) =
3
3
2
b2 t 1 + b1 t 2
The frictional force acting on the slider per unit width (drag force) is given by
U t dp
3
( - )
dx = (0b +b ) + dx = U b1 + b2 + b1 b32 t 1 t 32
b2 t 1 + b1 t 2
t 2 dx
t1 t 2
( +b2 )
F = 0b1
Condition for the maximum load:
Let b1+b2 = b = constant and x = t1/t2 and y = b1/b2. Then W = [3Ub2/t12]y(x - 1)/[(1+y)(y+x3)]. For a given
t2 , W is maximum when W/y = 0 and W/x = 0. These two conditions simplify to x3 = y2 and y = (3-2x)x2.
And the solutions are x=1.866; y=2.549. Then Wmax = 0.206U(b/t2)2.
Journal Bearings
Journal bearing is the most common type of hydrodynamic bearing in
use. It consists of a circular shaft (or Journal) rotating inside a circular
bush. If the journal and the bush are not co-axial, the bearing can carry a
transverse load.
h+e
r
e
h-e
Eccentricity Ratio = e/h
In general 0.4 < < 0.8
Note: Small may result in instabilities of the bearing operation.
Large may result in some degree of metal to metal contact (surface irregularity problems).
Clearance ratio = h/r = (R - r)/r
In general 110-3 << 410-3
Note: If is small, then h is small and it is necessary to machine the surfaces properly.
If is large, the rate of flow of the lubricant is high
Slenderness ratio = b/2r , where b is the width. In general 0.5 < <1.0
Note: In turbines b/2r0.8
Pressure Distribution:
x = r
t = ecos + Rcos - r ecos + R - r = h + ecos
Let at t = t0 , dp/dx = 0.
Then for the continuity (r)t/2 + (t3/12)(-dp/dx) = (r)t0/2 .
Therefore
= h(1 + cos)
R
e
dp/dx = 6(r)(t - t0)/t3 and dp/d = 6(r2)(t - t0)/t3 , where dx = rd, t = h(1 + cos).
Above differential equation, with the boundary condition [p] = [p]+2 , can be integrated to find an
expression for the pressure distribution. The result is
p h2
6 r
= Cp =
sin (2 + cos )
+ C0
(2 + 2 )(1 + cos )2
This was first obtained by Ernold Sommerfeld (1868-1951).
CP - C0
CP - C0
The existence of these symmetrical +ve and -ve
sections of the pressure distribution is termed the
Summerfeld boundary condition. In practice, usually
this condition can not be achieved. This is due to the
breakdown of lubricant film as the pressure falls below
the vapour pressure of the lubricant. In this situation
only the +ve section of the pressure distribution exists.
This is known as half-Summerfeld boundary condition.
A continuous film of lubricant can be preserved if the
value of C0 is sufficiently large. This can be achieved
by introducing the lubricant under pressure through a
small hole at some point in the bush and eventually
leaves via the ends of the bearing
Frictional torque/unit width T = r0(rd) = 4r3(1+22)/[h(1-2)1/2(2+2)]
Load Carrying Capacity
- Along = 0;
F =(0 sin - p cos)(rd) = 0
- Along = /2;
P =(p sin +0 cos)(rd) = 12r3/[h2(1-2)1/2(2+2)]
at Full-Summerfeld Condition.
Note: contribution of 0 is of order h/r times that of p and therefore neglected.
